3 Answers

If a charged particle is moving perpendicular to a magnetic field, it will start curving. It will go around a circle and come back to where it started. The radius of that circle will be r.

You can change the magnetic force by changing B or v or maybe even q. You might think, then it will no longer be equal to mv^2/r. But r will change so that mv^2/r changes to match Bqv.

Imagine you increase the magnetic field. Then the charged particle will make a tighter turn. If you decrease it, it makes a wider turn. Either way, it keeps turning the same direction, and eventually comes back to where it started.

If it moves in a circle at a constant speed, then Bqv = mv^2/r must be true for some value of r.

Why does it move in a circle at a constant speed? The assumptions are that there are no other forces on the particle and that the B field is uniform. The magnetic force is always perpendicular to the direction of motion. A perpendicular force can do no work, so it can't add or remove kinetic energy. That explains the constant speed. When something moves in a circle, the direction of motion, which is tangent to the circle, is perpendicular to the acceleration, which points to the center along the radius of the circle. That explains why a circle is a path that can result from a perpendicular force.

If mass greater, centripetal tension will advance because of the fact F=ma. If radius greater, centripetal tension decreases because of the fact centripetal acceleration, a.ok.a. radial acceleration is a=v^2/r. Plug that into F=ma, and you get F=(mv^2)/r. If tangential velocity greater, centripetal tension might advance.